Question

A playground is on the flat roof of a city school, 4.9 m above the street below (see figure). The vertical wall of the building is h = 6.00 m high, forming a 1.1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of ? = 53.0

Answer 1

Since the returned ball takes 2.2s. to its max. height, it reaches a height of 1/2 (t^2 x g) = 23.72 metres.
Initial vertical V component = (gt) = 9.8 x 2.2 = 21.56m/sec.
The ball also takes 2.2s. to reach the wall, so horizontal V component = (24/2.2) = 10.91m/sec.
a) Launch speed = sqrt.(10.91^2 + 21.56^2) = 24.16m/sec.
b) (23.72 - 6) = 17.72 metres.
c) The ball has to drop vertically (17.72 + 1.1) = 18.82 metres to the roof.
Time to drop = sqrt.(2h/g) = 1.96 secs.
In that time, it has advanced (1.96 x 10.91) = 21.38 metres horizontally, from the wall. at point b of beam ab determine graphically the magnitude and direction of their resultant using